A ball is released from the top of a tower of height `H m`. After `2 s` is stopped and then instantaneously released. What will be its heitht after next `2 s`?.

Options

A. `(H-5) m`
B. ` (H-10) m`
C. ` (H-20) m`
D. `(H-40) m`

Explanation

To solve the problem step by step, we will analyze the motion of the ball in two parts: the first 2 seconds when it is falling, and the next 2 seconds after it is stopped and then released again.

### Step 1: Calculate the distance fallen in the first 2 seconds
When the ball is released from the top of the tower, it falls freely under the influence of gravity. The initial velocity (u) is 0 m/s, and the acceleration (a) due to gravity is approximately 10 m/s².

Using the second equation of motion:
\[ S = ut + \frac{1}{2} a t^2 \]

Where:
- \( S \) = distance fallen
- \( u \) = initial velocity = 0 m/s
- \( a \) = acceleration = 10 m/s²
- \( t \) = time = 2 s

Substituting the values:
\[ S = 0 \cdot 2 + \frac{1}{2} \cdot 10 \cdot (2^2) \]
\[ S = 0 + \frac{1}{2} \cdot 10 \cdot 4 \]
\[ S = 20 \text{ meters} \]

### Step 2: Calculate the distance fallen in the next 2 seconds
After 2 seconds, the ball is stopped and then instantaneously released again. When it is released again, it starts from rest (initial velocity = 0 m/s) and falls for another 2 seconds under the influence of gravity.

Using the same equation of motion:
\[ S = ut + \frac{1}{2} a t^2 \]

Where:
- \( u = 0 \) m/s (initial velocity after being released)
- \( a = 10 \) m/s²
- \( t = 2 \) s

Substituting the values:
\[ S = 0 \cdot 2 + \frac{1}{2} \cdot 10 \cdot (2^2) \]
\[ S = 0 + \frac{1}{2} \cdot 10 \cdot 4 \]
\[ S = 20 \text{ meters} \]

### Step 3: Calculate the total distance fallen
Now, we need to find the total distance fallen by the ball after the first 4 seconds (2 seconds + 2 seconds):
Total distance fallen = Distance in first 2 seconds + Distance in next 2 seconds
\[ \text{Total distance} = 20 \text{ m} + 20 \text{ m} = 40 \text{ meters} \]

### Step 4: Calculate the height of the ball after 4 seconds
The height of the ball after falling 40 meters from the top of the tower (height \( H \)):
\[ \text{Height after 4 seconds} = H - 40 \text{ meters} \]

### Final Answer
The height of the ball after the next 2 seconds is:
\[ H - 40 \text{ meters} \]